Question

JAN 2019
BAY
14 28 M
15 29
Itu
WK OO
It a 21 16 30 w
t
I
L
19
dot
31 17 31 T
4 18
F
5) 19 5
20 S
It a
ltn
Itn
Hn
OM 121
T 8 22.
W 9 23
T 10 24​

Answer 1

Answer:

This implies that

x2+2ax=4x−4a−13

or

x2+2ax−4x+4a+13=0

or

x2+(2a−4)x+(4a+13)=0

Since the equation has just one solution instead of the usual two distinct solutions, then the two solutions must be same i.e. discriminant = 0.

Hence we get that

(2a−4)2=4⋅1⋅(4a+13)

or

4a2−16a+16=16a+52

or

4a2−32a−36=0

or

a2−8a−9=0

or

(a−9)(a+1)=0

So the values of a are −1 and 9.

achaa